Математичнi Студiї. Т.37, №1
Matematychni Studii. V.37, No.1
УДК 517.53
M. P. Mahola, P. V. Filevych
THE ANGULAR VALUE DISTRIBUTION OF RANDOM
ANALYTIC FUNCTIONS
M. P. Mahola, P. V. Filevych. The angular value distribution of random analytic functions,
Mat. Stud. 37 (2012), 34–51.
P
Let R ∈ (0, +∞], f (z) =
cn z n be an analytic function in the disk {z : |z| < R}, Tf (r)
be the Nevanlinna characteristic, Nf (r, α, β, a) be the integrated counting function of a-points
of f in the sector 0 < |z| ≤ r, α ≤ argα z < β, and (ωn (ω)) be a sequence of independent
equidistributed on [0, 1] random variables. Under some
conditions on the growth of f it is
P 2πiω
n (ω)
proved that for random analytic function fω (z) =
e
an z n almost surely for every
β−α
a ∈ C and all α < β ≤ α + 2π the relation Nfω (r, α, β, a) ∼ 2π Tfω (r), r → R, holds outside
some exceptional set E ⊂ (0, R).
М. П. Магола, П. В. Филевич. Угловое распределение значений случайных аналитических
функций // Мат. Студiї. – 2012. – Т.37, №1. – C.34–51.
P
Пусть R ∈ (0, +∞], f (z) =
cn z n — аналитическая в круге {z : |z| < R} функция,
Tf (r) — характеристика Неванлинны, Nf (r, α, β, a) — усредненная считающая функция
a-точек функции f в секторе 0 < |z| ≤ r, α ≤ argα z < β, а (ωn (ω)) — последовательность независимых равномерно распределенных на [0, 1] случайных величин. При
некоторых
P условиях на рост f доказано, что для случайной аналитической функции
fω (z) = e2πiωn (ω) an z n почти наверное для всех a ∈ C и любых α < β ≤ α + 2π вне некоторого исключительного множества E ⊂ (0, R) выполняется соотношение Nfω (r, α, β, a) ∼
β−α
2π Tfω (r), r → R.
1. Introduction. Let D(r) = {z ∈ C : |z| < r} for all r ∈ (0, +∞], ln+ x = ln max{x, 1} for
each x ∈ [0, +∞), and S(r, α, β) = {z ∈ C : 0 < |z| ≤ r, α ≤ argα z < β} for any α, β ∈ R
such that α < β ≤ α + 2π (here, for a complex number z 6= 0, argα z is the value of its
argument, which belongs to the interval [α, α + 2π)). By L we denote the class of positive
unbounded nondecreasing functions on [0, +∞).
We consider a measurable set E ⊂ R, and let R ∈ (0, +∞]. As usual, if R = +∞
(R < +∞), then the integral
Z
Z
dr
dr
E∩(0,R) R − r
E∩(1,+∞) r
is called the logarithmic measure of the set E on (0, R). The limits
Z
Z
Z
Z
dt
dt
(R − r)dt
(R − r)dt
lim
, lim
lim
, lim
2
r→+∞ E∩(0,r) r
r→R E∩(0,r) (R − t)2
r→+∞ E∩(0,r) r
r→R E∩(0,r) (R − t)
2010 Mathematics Subject Classification: 30B20, 30D20, 30D35.
Keywords: meromorphic function, value distribution, random analytic function, Nevanlinna characteristic,
counting function, integrated counting function.
c M. P. Mahola, P. V. Filevych, 2012
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
35
are called the upper density and the lower density of the set E on (0, R), respectively.
We say that a set E has density d on (0, R), if its upper density and its lower density on
(0, R) are equal to d. It is easy to prove that every set E of finite logarithmic measure on
(0, R) has density 0 on (0, R).
All functions meromorphic (in particular, analytic) in a disk considered below are assumed
to be different from constants.
We use the standard notations from the value distribution theory of meromorphic functions ([1, 2]). In particular, if R ∈ (0, +∞], r ∈ (0, R), α < β ≤ α + 2π, and f is a function
meromorphic in D(R), then let nf (r) be the counting functions of poles of the function f ,
n
ef (r) = nf (r) − nf (0), and n
ef (r, α, β) be the counting functions of poles of the function f in
the sector S(r, α, β). We define the integrated counting functions of poles, integrated counting
functions of poles in the sector S(r, α, β), proximity function, Nevanlinna characteristic, and
maximum modulus of the function f by
Z r
Z r
dt β − α
dt
n
ef (t, α, β) +
nf (0) ln r,
Nf (r) =
n
ef (t) + nf (0) ln r, Nf (r, α, β) =
t
t
2π
0
0
Z 2π
1
mf (r) =
ln+ |f (reiθ )|dθ, Tf (r) = Nf (r) + mf (r), Mf (r) = sup{|f (z)| : |z| = r},
2π 0
respectively. For every a ∈ C we put Xf (r, a):=X 1 (r), where X is some of the characterif −a
stics n, n
e, N , m or T , n
ef (r, α, β, a) = n
e 1 (r, α, β), Nf (r, α, β, a) = N 1 (r, α, β), and let
f −a
f −a
cf (a) be the first non-zero coefficient in the Laurent series of the function f (z) − a in a neighborhood of the point z = 0.
Denote by H(R) the class of all functions analytic in the disk D(R) of the form
f (z) =
∞
X
cn z n
(1)
n=0
1
P
2 2n 2
such that Sf (r):= ( ∞
n=0 |cn | r ) → +∞ (r → R).
Consider a probability space (Ω, A, P ), where Ω is some set, A is a σ-algebra of subset
of Ω, P is a complete probability measure on (Ω, A), and suppose that on this space there
exists a Steinhaus sequence (ωn (ω)), i. e. a sequence of independent uniformly distributed
on [0, 1] random variables (see [3]). From now on we assume that such a probabilistic space
and a corresponding Steinhaus sequence are given and fixed.
Along with an analytic function f ∈ H(R) of the form (1) we consider the random
analytic function
∞
X
fω (z) =
e2πiωn (ω) cn z n .
(2)
n=0
The value distribution of random analytic functions of form (2) were studied in the papers
[4] (for R = 1) and [5] (for R = +∞). In particular, in [5] it is proved the following theorems
(Theorem A is proved for R = +∞).
Theorem A. Let R ∈ (0, +∞], and f ∈ H(R) be an analytic function of form (1). Then
for the random analytic function defined by (2) almost surely (a. s.) the inequality
ln Sf (r) ≤ Nfω (r, 0) + C0 ln Nfω (r, 0) (r0 (ω) ≤ r < R)
holds, where C0 > 0 is an absolute constant.
36
M. P. MAHOLA, P. V. FILEVYCH
R +∞ dx
< +∞. Then
Theorem B. Let f be an entire function of form (1), ϕ ∈ L, and 0 ϕ(x)
there exists a set E of finite logarithmic measure on (0, +∞) such that for the random entire
function defined by (2) a. s. for every a ∈ C we have
ln Sf (r) ≤ Nfω (r, a) + ln2 Nfω (r, a)ϕ(ln Nfω (r, a)) (r ≥ r0 (ω, a), r ∈
/ E).
The proof of Theorem A for the case R ∈ (0, +∞) is analogous to that for the case
R = +∞ given in [5]. So, we assume that Theorem A is proved for all R ∈ (0, +∞].
In this paper we consider some problems concerning the angular value distribution of
random analytic functions of form (2). We also use some refinements to make Theorem B
more precise.
Note that questions about the angular value distribution of analytic functions in the
terms of characteristic Nf (r, α, β, a) were investigated in [6]–[8]. Mainly these papers deal
with entire functions (in particular, entire functions presented by lacunary power series),
satisfying the condition
ln Mf (r) ∼ Tf (r) (E1 3 r → +∞),
(3)
where E1 is a set, that is large in some sense. The following result of W. K. Hayman and
J. F. Rossi [8] is one of the most general in this direction.
Theorem C. Let f be an entire function of the order
ln ln Mf (r)
>0
r→+∞
ln r
ρf := lim
such that relation (3) holds on a set E1 of density 1 on (0, +∞). Then there exists a set E2
of upper density 1 on (0, +∞) such that for every a ∈ C and all α, β ∈ R, α < β ≤ α + 2π,
we have
β−α
Tf (r) (E2 3 r → +∞).
Nf (r, α, β, a) ∼
2π
The next assertion follows from Theorems A and C.
Corollary A. Let f be an entire function of the order ρf > 0 and form (1). Then for the
random entire function defined by (2) a. s. there exists a set Eω of upper density 1 on (0, +∞)
such that
β−α
Nfω (r, α, β, a) ∼
ln Sf (r)
(4)
2π
as Eω 3 r → +∞ for every a ∈ C and all α, β ∈ R, α < β ≤ α + 2π.
We omit a justification of Corollary A, since below we shall prove a stronger statement.
The following theorems are the main results of our paper.
Theorem 1. Let R ∈ (0, +∞], and let f ∈ H(R) be an analytic function of form (1). Then
there exists a function h ∈ L such that for the random analytic function defined by (2) a. s.
for every a ∈ C the inequality
ln Sf (r) ≤ Nfω (r, a) + C1 ln ln Sf (R) + ln
holds, where C1 > 0 is an absolute constant.
R
+ h(|a|) (r1 (ω) ≤ r < R < R),
R−r
(5)
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
37
Theorem 2. Let R ∈ (0, +∞], let f ∈ H(R) be an analytic function
of form (1), let
p
r0 ∈ (0, R) be an arbitrary fixed number such that Sf (r0 ) ≥ max{e, 1 + |c0 |2 }, and
R
lf (r) = min ln ln Sf (R) + ln
: R ∈ [r, R)
(r0 < r < R).
R−r
Then for the random analytic function defined by (2) a. s. for every a ∈ C there exists
a constant C = C(ω, a) > 0 such that for all α, β ∈ R, α < β ≤ α + 2π, we have
1
Z r
r dt 3
β−α
2
(lf (t) + C) ln
ln Sf (r) + C2 ln (Sf (r) + C)
+C
Nfω (r, α, β, a) ≤
2π
t t
r0
for each r ∈ (r0 , R), where C2 > 0 is an absolute constant.
Next, we formulate some corollaries from Theorems 1 and 2.
Corollary 1. Let f be an entire function of form (1). Then there exist a function h ∈ L and
a set E3 of finite logarithmic measure on (0, +∞) such that for the random entire function
defined by (2) a. s. for every a ∈ C the inequality
/ E3 )
ln Sf (r) ≤ Nfω (r, a) + C3 ln ln Sf (r) + h(|a|) (r ≥ r2 (ω), r ∈
(6)
holds, where C3 > 0 is an absolute constant.
Corollary 2. Let f be an entire function of form (1) such that
ln Sf (r)
= +∞.
r→+∞ ln2 r ln ln r
lim
(7)
Then there exists a set E4 of upper density 1 on (0, +∞) such that for the random entire
function defined by (2) a. s. for every a ∈ C and all α, β ∈ R, α < β ≤ α + 2π, relation (4)
holds as E4 3 r → +∞.
Corollary 3. Let ρ ∈ (0, +∞), and let f be an entire function of the order ρf ≥ ρ and
form (1). Then there exists a set E5 of upper density 1 on (0, +∞) such that for the random
entire function defined by (2) a. s. for every a ∈ C and all α, β ∈ R, α < β ≤ α + 2π, the
inequality
2
C4
Nfω (r, α, β, a) − β − α ln Sf (r) ≤ p
ln 3 Sf (r) ln ln Sf (r) (r ≥ r3 (ω, a), r ∈ E5 ) (8)
3
2π
ρ2
holds, where C4 > 0 is an absolute constant.
Corollary 4. Let f be an entire function of finite order and form (1). Then for the random
entire function defined by (2) a. s. for every a ∈ C we have
Nfω (r, a) ∼ ln Sf (r) (r → +∞).
(9)
Corollary 5. Let f be an entire function of finite order and form (1) such that
ln Sf (r)
= +∞.
r→+∞ ln3 r
lim
(10)
Then for the random entire function defined by (2) a. s. for every a ∈ C and all α, β ∈ R,
α < β ≤ α + 2π, relation (4) holds as r → +∞.
38
M. P. MAHOLA, P. V. FILEVYCH
Corollary 6. Let R ∈ (0, +∞), and let f ∈ H(R) be an analytic function of form (1) such
that
ln Sf (r)
lim
= +∞.
(11)
r→R ln 1
R−r
Then there exists a set E6 of upper density 1 on (0, R) such that for the random analytic
function defined by (2) a. s. for every a ∈ C and all α, β ∈ R, α < β ≤ α + 2π, relation (4)
holds as E6 3 r → R.
Concluding the Introduction, we note that the value distribution and other properties of
some classes of random analytic functions were investigated also in [9]–[24].
2. Auxiliary results. Let x1 , . . . , xn ∈ [0, +∞). The following inequalities
n
n
n
n
X
X
X
Y
+
+
+
xν ≤
ln+ |xn | + ln n
ln |xn |, ln ln xν ≤
ν=1 ν=1
ν=1 ν=1
are well known (see, for example, [2], p. 14). Below we will use these inequalities without
additional explanations.
The following lemma is proved in [25].
Lemma A. Let R ∈ (0, +∞], and let g be a meromorphic function in the disk D(R) such
that g(0) = 1. Then for arbitrary α, β ∈ (0, 1) the inequality
α
Z 2π 0 iθ α
g (re ) 1
dθ ≤ C(α, β) Tg (R) R
(0 < r < R < R)
(12)
2π 0 g(reiθ ) r R−r
is true, where

C(α, β) =
2
1−β
α
4 + 2
+
1+α
1−α
+2
βα
2+α
1−α
1−α 
απ

.
 sec
2
For a function f meromorphic in D(R) and every z ∈ D(R) we put g ∗ (z) = zg 0 (z). Then
inequality (12) is equivalent to the inequality
α
Z 2π ∗ iθ α
g (re ) 1
R
dθ ≤ C(α, β) Tg (R)
(0 < r < R < R).
(13)
2π 0 g(reiθ ) R−r
Arguing as in the paper [26] in the proof of its main result, and using inequality (13)
instead of inequality (12), it is easy to prove the following statement.
Lemma B. Let R ∈ (0, +∞], and let g be a function meromorphic in the disk D(R) such
that g(0) = 1. Then
R
+
m g∗ (r) ≤ ln Tg (R)
+ 4, 8517 (0 < r < R < R).
g
R−r
Lemma 1. Let F ⊂ [0, 2π] be a measurable set, R ∈ (0, +∞], r ∈ (0, R), f be a function
analytic in the disk D(R) of form (1). Then
Z
1
1
µ(F) +
ln+ |f (reiθ )|dθ ≤
+
ln Sf (r),
(14)
2π F
2e
2π
where µ(F) is the Lebesgue measure of the set F.
39
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
Proof. Let E = {θ ∈ F : |f (reiθ )| > 1}. If µ(E) = 0 then inequality (14) is trivial. If µ(E) > 0,
then, using the Jensen inequality (see, for example, [27], p. 42)
Z
Z
1
1
iθ 2
iθ 2
ln |f (re )| dθ ≤ ln
|f (re )| dθ
µ(E) E
µ(E) E
and the Parseval equality
Z
2π
|f (reiθ )|2 dθ = 2πSf2 (r),
0
we obtain
Z
Z
Z
1
1
µ(E)
1
+
iθ
iθ 2
iθ 2
ln |f (re )|dθ =
ln |f (re )| dθ ≤
ln
|f (re )| dθ ≤
2π F
4π E
4π
µ(E) E
Z 2π
µ(E)
1
µ(E)
2π
µ(E)
iθ 2
≤
ln
|f (re )| dθ =
ln
+
ln Sf (r).
4π
µ(E) 0
4π
µ(E)
2π
Since the most value of the function y(x) =
Lemma 1 is proved.
x
2
ln x1 on the interval (0, +∞) is equal to
1
,
2e
Lemma 2. Let R ∈ (0, +∞], and let f be a function analytic in the disk D(R). Then for
every a ∈ C and all r, R ∈ (0, R), r < R, we have
m
f∗
f −a
(r) ≤ ln+ ln+ Sf (R) + ln
R
1
+ ln+
+
R−r
|cf (a)|
1
+ ln+ nf (0, a) + ln+ |a| + 7.
R
Proof. We fix arbitrary a ∈ C and r, R ∈ (0, R), r < R. Put
+nf (0, a) ln+
g(z) =
f (z) − a
cf (a)z nf (0,a)
(15)
(z ∈ D(R)).
It is easily verified that
f ∗ (z)
g ∗ (z)
=
+ nf (0, a) (z ∈ D(R)).
f (z) − a
g(z)
Consequently,
m
f∗
f −a
(r) ≤ m g∗ (r) + ln+ nf (0, a) + ln 2.
(16)
g
In addition, g(0) = 1. Therefore, by Lemma B, we have
m g∗ (r) ≤ ln+ Tg (R) + ln
g
R
+ 4, 8517.
R−r
Next note that Lemma 1 implies the inequality
1
Tf (r) ≤
+ ln+ Sf (r) (r ∈ (0, R)).
2e
Using this inequality with R instead of r, we obtain
1
1
ln+ Tg (R) ≤ ln+ Tf (R) + ln+ |a| + ln 2 + ln+
+ nf (0, a) ln+ ≤
|cf (a)|
R
1
1
≤ ln+ ln+ Sf (R) + 2 ln 2 + ln+ |a| + ln+
+ nf (0, a) ln+ .
|cf (a)|
R
Then inequality (15) is an obvious consequence from inequalities (16), (17), and (19).
(17)
(18)
(19)
40
M. P. MAHOLA, P. V. FILEVYCH
Let R ∈ (0, +∞], r ∈ (0, R), and let g be a function analytic in the disk D(R). We set
Eg (r) = {θ ∈ R : g(teiθ ) 6= 0 for all t ∈ (0, r]}.
Note, that Eg (r2 ) ⊂ Eg (r1 ) if 0 < r1 < r2 < R. The set Eg (r) is periodic in the sense that
θ ∈ Eg (r) if and only if (θ + 2π) ∈ Eg (r). In addition, [0, 2π)\Eg (r) is a finite set for all
r ∈ (0, R).
Suppose that g(0) = 1, and fix an arbitrary θ ∈ Eg (r). Then g(teiθ ) 6= 0 for each t ∈ [0, r].
In view of this, by vg (t, θ) we denote the continuous branch of the argument of the function
g(teiθ ) such that vg (0, θ) = 0, and put
Z r
dt
1
vg (t, θ) .
Vg (r, θ) =
2π 0
t
The following statement is well known (see [8], [6], and [28], p. 126).
Lemma C. Let R ∈ (0, +∞], r ∈ (0, R), and let g be a function analytic in the disk D(R)
such that g(0) = 1. Then:
(i) for all α, β ∈ Eg (r) such that α < β ≤ α + 2π we have
Z β
1
Ng (r, α, β, 0) =
ln |g(reiθ )|dθ + Vg (r, α) − Vg (r, β);
2π α
(ii) for all α, β ∈ R such that α < β ≤ α + 2π we have
Z β
Z r
1
r dt
Vg (r, θ)dθ =
(ln |g(teiα )| − ln |g(teiβ )|) ln
.
2π 0
t t
α
Let g be a function analytic in the disk D(R) such that g(0) = 1. Consider an arbitrary
interval (ϕ, ψ) ⊂ Eg (r), fix some point α in this interval, and let β 6= α be an arbitrary point
of this interval. Then the function g has no zeros in the sector S(r, min{α, β}, max{α, β}).
Therefore, Ng (r, min{α, β}, max{α, β}, 0) = 0. According to (i) of Lemma C we have
Z β
1
ln |g(reiθ )|dθ.
Vg (r, β) = Vg (r, α) +
2π α
Rβ
Since for a fixed α the function y(β) = α ln |g(reiθ )|dθ is continuous and bounded on every
finite interval of the real axis, then Vg (r, β), as a function of β, is continuous and bounded
on the interval (ϕ, ψ). From the above considerations, as well as from the periodicity of the
set Eg (r) and the finiteness of the set [0, 2π)\Eg (r), we obtain that the function Vg (r, β) is
continuous and bounded on Eg (r).
Now let f be an arbitrary function analytic in the disk D(R). Put
g(z) =
f (z)
.
cf (0)z nf (0,0)
Then g(0) = 1, Ef (r) = Eg (r), and n
ef (r, α, β, 0) = n
eg (r, α, β, 0). Therefore, setting Vf (r, θ) =
Vg (r, θ) for all θ ∈ Ef (r), from Lemma C, as a consequence, we obtain the following statement.
Lemma D. Let R ∈ (0, +∞], r ∈ (0, R), and let f be a function analytic in the disk D(R).
Then there exists a function Vf (r, θ) continuous and bounded on Ef (r) such that:
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
41
(i) for all α, β ∈ Ef (r) such that α < β ≤ α + 2π we have
Z β
β−α
1
ln |f (reiθ )|dθ −
ln |cf (0)| + Vf (r, α) − Vf (r, β);
Nf (r, α, β, 0) =
2π α
2π
(ii) for all α, β ∈ R such that α < β ≤ α + 2π we have
Z β
Z r
1
r dt
Vf (r, θ)dθ =
.
(ln |f (teiα )| − ln |f (teiβ )|) ln
2π 0
t t
α
Note that the equality from assertion (i) of Lemma D is a generalization of the following
classical Jensen equality
Z 2π
1
Nf (r, 0) =
ln |f (reiθ )|dθ − ln |cf (0)| (r ∈ (0, R)).
(20)
2π 0
Also note that equality (20) implies the inequality
Nf (r, 0) ≤ Tf (r) − ln |cf (0)| (r ∈ (0, R)).
(21)
In fact, the following statement [29] is an immediate consequence of the classical BorelNevanlinna lemma (see [2], p. 90).
Lemma E. Let u(r) be a nondecreasing function unbounded on [r0 , +∞), x0 = u(r0 ),
and
ϕ(x) be a continuous positive function increasing to +∞ on [x0 , +∞) such that
R +∞ let
dx
< +∞. Then for all r ≥ r0 outside a set E of finite logarithmic measure on
ϕ(x)
x0
(0, +∞) we have
1
u r exp
< eu(r).
ϕ(ln u(r))
Lemma 3. Let 0 ≤ r0 < R < +∞, u(r) be a nondecreasing function unbounded on [r0 , R),
x0 = u(r0 ),R and let ϕ(x) be a continuous positive function increasing to +∞ on [x0 , +∞)
+∞ dx
< +∞. Then for all r ∈ [r0 , R) outside a set E of finite logarithmic
such that x0 ϕ(x)
measure on (0, R) we have
1
u R − (R − r) exp −
< eu(r).
ϕ(ln u(r))
Proof. It suffices to show that the set
F = r ∈ [r0 , R) : u R − (R − r) exp −
1
ϕ(ln u(r))
≥ eu(r)
has finite logarithmic measure on (0, R).
Put
0
R
R
r −1
0
0
0
r =
, r0 =
, v(r ) = u R 0
.
R−r
R − r0
r
Then r = R 1 − r10 . It is easy to verify that the set F is the image of the set
1
0
0
0
0
0
F = r ≥ r0 : v r exp
≥ ev(r )
ϕ(ln v(r0 ))
42
M. P. MAHOLA, P. V. FILEVYCH
under the mapping r = R 1 − r10 . By Lemma E, the set F 0 has finite logarithmic measure
on (0, +∞). Therefore,
Z
Z
Z
dr
r0
dr0
1
=
dR 1 − 0 =
< +∞,
0
r
F R−r
F0 R
F0 r
i. e. the set F has finite logarithmic measure on (0, R).
3. Proofs of theorems and corollaries.
Proof of Theorem 1. Let R ∈ (0, +∞], f ∈ H(R) be an analytic function of form (1), and
m = min{n ∈ N : cn 6= 0}. Fix an arbitrary r0 ∈ (0, R) such that Sf ∗ (r) ≥ Sf (r) ≥ e
(r ≥ r0 ), and for each x ∈ [0, +∞) put
h0 (x) = m ln+
1
+ ln+ m + ln+ x + 7,
r0
h(x) = h0 (x) + ln+ max{|c0 | + x, |cm |} + ln+
1
.
m|cm |
It is clear that h ∈ L.
Consider the random analytic function defined by (2). Then, as easily seen, for all ω ∈ Ω
and each a ∈ C the relations
|cfω∗ (0)| = m|cm |,
|cfω (a)| ≤ max{|c0 | + |a|, |cm |},
|nfω (0, a)| ≤ m
are true.
Let C0 be the constant from Theorem A, and A is the following event: there exists
r0 (ω) ∈ (0, R) such that
ln Sf ∗ (r) ≤ Nfω∗ (r, 0) + (C0 + 1) ln ln Sf ∗ (r) (r0 (ω) ≤ r < R).
(22)
By Theorem A we have P (A) = 1. Furthermore, since Sf ∗ (r) ≥ Sf (r) (r ≥ r0 ) and the
function y(x) = x − (C0 + 1) ln x is increasing on [x0 , +∞), for every ω ∈ A satisfying (22)
we obtain
ln Sf (r) ≤ Nfω∗ (r, 0) + (C0 + 1) ln ln Sf (r) (r1 (ω) ≤ r < R),
(23)
where r1 (ω) ≥ r0 .
Fix an arbitrary ω ∈ Ω. Using Jensen’s formula (20), written for the functions fω∗ and
fω − a, and Lemma 2 for the function fω instead of f , for each a ∈ C and all r, R ∈ [r0 , R),
r < R, we have
Z 2π fω∗ (reiθ ) 1
dθ − ln |cfω∗ (0)| ≤
Nfω∗ (r, 0) − Nfω (r, a) =
ln 2π 0
fω (reiθ ) − a |cfω (a)|
∗
|cf (0)|
|cf ∗ (0)|
R
1
≤ m fω∗ (r) − ln ω
≤ ln ln Sf (R) + ln
+ ln+
+ h0 (|a|) − ln ω
=
fω −a
|cfω (a)|
R−r
|cfω (a)|
|cfω (a)|
R
= ln ln Sf (R) + ln
+ h0 (|a|) + ln+ |cfω (a)| − ln |cfω∗ (0)| ≤
R−r
R
≤ ln ln Sf (R) + ln
+ h(|a|).
R−r
From this and from (23) for arbitrary ω ∈ A and a ∈ C we obtain
ln Sf (r) ≤ Nfω (r, a) + (C0 + 2) ln ln Sf (R) + ln
R
+ h(|a|) (r1 (ω) ≤ r < R < R).
R−r
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
43
Theorem 2 is obtained from Theorem 1 and the following statement.
Theorem 3. Let R ∈ (0, +∞], g ∈ H(R), and let r0 ∈ (0, R) be an arbitrary fixed number
such that Sg (r0 ) ≥ 1, and
Z r
1
r dt
ln Sg (t) − Ng (t, 0) − ln |cg (0)| +
ln
(r0 < r < R).
(24)
hg (r) =
e
t t
r0
Then there exists a constant C7 > 0 such that for all α, β ∈ R, α < β ≤ α + 2π, we have
r
3 2
β−α
3
Ng (r, α, β, 0) ≤
ln Sg (r) + 3
ln Sg (r)hg (r) + C7 (r0 < r < R).
(25)
2π
π2
Proof. Let C7 = 1e + 2C8 + 2| ln |cg (0)||, where C8 is a constant such that |Vg (r0 , θ)| ≤ C8 for
every θ ∈ Eg (r0 ) (see Lemma D).
Fix arbitrary r > r0 and α, β ∈ R such that α < β ≤ α + 2π. If 8hg (r) ≥ ln Sg (r), then
inequality (25) holds. Indeed, using inequalities (21) and (18) with the function g instead
of f , we have
β−α
ln Sg (r) + Ng (r, 0) ≤
2π
q
β−α
1
β−α
≤
ln Sg (r) + ln Sg (r) +
− ln |cg (0)| ≤
ln Sg (r) + 2 3 ln2 Sg (r)hg (r) + C7 .
2π
2e
2π
Ng (r, α, β, 0) ≤ Ng (r, 0) ≤
Now let 8hg (r) < ln Sg (r) and
s
ε=
3
8π hg (r)
.
9 ln Sg (r)
1
Inequalities (21) and (18) imply that ln Sg (t) − Ng (t, 0) − ln |cg (0)| + 1e ≥ 2e
for all t ∈ (r0 , r).
Thus, hg (r) > 0. Moreover, ln Sg (r) > 0. Therefore, ε > 0. On the other hand,
r
π
3 8π 1
ε<
< .
9 8
4
Put
1
ϕ(θ) =
2π
Z
r
r0
ln |g(teiθ )| ln
r dt
.
t t
Then, applying Lemma 1 to the function g, we obtain
Z α−2ε
Z r Z α−2ε
Z r
r dt
1
1
ε
r dt
iθ
I1 :=
ϕ(θ)dθ =
ln |g(te )|dθ ln
≤
+
ln Sg (t) ln
.
2π α−3ε
t t
2e 2π
t t
α−3ε
r0
r0
From this and from the mean value theorem, applied to the integral I1 , it follows the existence
of a number ζ1 ∈ [α − 3ε, α − 2ε] such that
Z 1 r 1
ε
r dt
ϕ(ζ1 ) ≤
+
ln Sg (t) ln
.
(26)
ε r0 2e 2π
t t
Since for any x, y ∈ R the Jensen formula implies the equality
Z y
Z x+2π
1
1
iθ
ln |g(re )|dθ = Ng (r, 0) + ln |cg (0)| −
ln |g(reiθ )|dθ,
2π x
2π y
x, y ∈ R,
44
M. P. MAHOLA, P. V. FILEVYCH
using Lemma 1, in the case x < y ≤ x + 2π we have
1
2π
y
Z
ln |g(reiθ )|dθ ≥ Ng (r, 0) + ln |cg (0)| −
x
1
x + 2π − y
−
ln Sg (r).
2e
2π
Then
Z
α
Z r
Z
α
r dt
≥
I2 :=
ϕ(θ)dθ =
ln |g(te )|dθ ln
t t
α−ε
r0
α−ε
Z r
1
2π − ε
r dt
Ng (t, 0) + ln |cg (0)| −
≥
−
ln Sg (t) ln
.
2e
2π
t t
r0
1
2π
iθ
Therefore, the mean value theorem applied to the integral I2 , yields the existence of a number
ζ2 ∈ [α − ε, α] such that
1
ϕ(ζ2 ) ≥
ε
Z r
2π − ε
1
r dt
−
ln Sg (t) ln
.
Ng (t, 0) + ln |cg (0)| −
2e
2π
t t
r0
(27)
Then, using Lemma D and inequalities (26) and (27), we obtain
Z r
r dt
1
(ln |g(teiζ1 )| − ln |g(teiζ2 )|) ln
=
I3 :=
(Vg (r, θ) − Vg (r0 , θ))dθ =
2π r0
t t
ζ1
Z
1 r
1
r dt
1
= ϕ(ζ1 ) − ϕ(ζ2 ) ≤
(ln Sg (t) − Ng (t, 0) − ln |cg (0)| + ) ln
= hg (r).
ε r0
e
t t
ε
Z
ζ2
From the inequality ζ2 − ζ1 ≥ ε and from the mean value theorem, applied to the integral I3 ,
it follows the existence of a number ζ ∈ [ζ1 , ζ2 ] ∩ Eg (r) such that
Vg (r, ζ) − Vg (r0 , ζ) ≤
1
hg (r).
ε2
(28)
Similarly we can prove that there exist numbers η1 ∈ [β, β + ε] and η2 ∈ [β + 2ε, β + 3ε]
such that
Z 1 r
1
2π − ε
r dt
Ng (t, 0) + ln |cg (0)| −
−
ln Sg (t) ln
,
ϕ(η1 ) ≥
ε r0
2e
2π
t t
Z ε
r dt
1 r 1
ϕ(η2 ) ≤
+
ln Sg (t) ln
.
ε r0 2e 2π
t t
Then
Z
η2
I4 :=
η1
1
(Vg (r, θ) − Vg (r0 , θ))dθ = ϕ(η1 ) − ϕ(η2 ) ≥ − hg (r).
ε
The inequality η2 − η1 ≤ 3ε together with the mean value theorem applied to the integral I4
implies the existence of a number η ∈ [η1 , η2 ] ∩ Eg (r) such that
Vg (r, η) − Vg (r0 , η) ≥ −
1
hg (r).
3ε2
(29)
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
45
Using Lemmas D and 1, as well as inequalities (28) and (29), and taking into account
that η − ζ ≤ β − α + 6ε < 4π, we obtain
Z η
η−ζ
1
ln |g(reiθ )|dθ −
ln |cg (0)| + Vg (r, ζ) − Vg (r, η) ≤
Ng (r, α, β, 0) ≤ Ng (r, ζ, η, 0) =
2π ζ
2π
η−ζ
1 η−ζ
≤
ln Sg (r) + −
ln |cg (0)| + Vg (r, ζ) − Vg (r, η) ≤
2π
e
2π
β − α + 6ε
1
1
1
≤
ln Sg (r) + + 2| ln |cg (0)|| + 2 hg (r) + C8 + 2 hg (r) + C8 =
2π
e
ε
3ε
r
3 2
β−α
3
ln Sg (r) + 3
=
ln Sg (r)hg (r) + C7 .
2π
π2
Proof of Theorem 2. Let f ∈ H(R) be an analytic function
p of form (1), and let r0 ∈ (0, R)
be an arbitrary fixed number such that Sf (r0 ) ≥ max{e, 1 + |c0 |2 }.
Consider the random analytic function defined by (2). For arbitrary ω ∈ Ω, a ∈ C, and
r ∈ (0, R) we have Sf2ω −a (r) = |c0 e2πiω0 (ω) − a|2 + Sf2 (r) − |c0 |2 . This implies the inequalities
Sf2 (r) − |c0 |2 ≤ Sf2ω −a (r) ≤ Sf2 (r) + (|c0 | + |a|)2 .
Then Sf2ω −a (r) ≥ 1 by the first of these inequalities. By the second of these inequalities, there
exists a constant C9 = C9 (a) > 0 such that
ln Sfω −a (r) ≤ ln Sf (r) + C9
(r0 < r < R).
(30)
Let C1 > 0 is the absolute constant and h ∈ L is the function, the existence of which
follows from Theorem 1. Let B be the next event: for every a ∈ C inequality (5) holds. Then,
by Theorem 1, P (B) = 1.
Fix arbitrary ω ∈ B and a ∈ C, and let g(z) = fω (z) − a. Then (5) and (30) implies the
existence of a constant C10 = C10 (ω, a) > 0 such that
ln Sg (r) − Ng (r, 0) − ln |cg (0)| +
1
≤ C1 (lf (r) + C10 ) (r0 < r < R).
e
So, if hg is the function defined by the equality (24), then
Z r
r dt
hg (r) ≤ C1
(lf (t) + C10 ) ln
(r0 < r < R).
t t
r0
(31)
By Theorem 3, there exists a constant C7 = C7 (ω, a) > 0 such that for arbitrary α, β ∈ R,
α < β ≤ α + 2π, inequality (25) holds. Using this inequality and also inequalities (30) and
(31), for arbitrary α, β ∈ R, α < β ≤ α + 2π, and all r ∈ (r0 , R) we obtain
r
3 2
β−α
3
ln Sg (r)hg (r) + C7 ≤
Nfω (r, α, β, a) = Ng (r, α, β, 0) ≤
ln Sg (r) + 3
2π
π2
1
Z r
β−α
3
r dt 3
2
ln Sf (r) + C9 + 3
C1 ln (Sf (r) + C9 )
(lf (t) + C10 ) ln
+ C7 .
≤
2π
π2
t t
r0
q
1
Finally, putting C(ω, a) = max{C9 (a) + C7 (ω, a), C10 (ω, a)} and C2 = 3 3 3C
, we complete
π2
the proof of Theorem 2.
46
M. P. MAHOLA, P. V. FILEVYCH
Proof of Corollary 1. Let f be an entire function of form (1), r0 = min{r ≥ 0 : ln Sf (r) ≥ e},
and x0 = ln Sf (r0 ). Put
1
(r ≥ r0 ).
R(r) = r exp
ln2 ln Sf (r)
Applying Lemma E to the functions u(r) = ln Sf (r) (r ≥ r0 ) and ϕ(x) = x2 (x ≥ x0 ), we
have
ln Sf (R(r)) < e ln Sf (r) (r ≥ r0 , r ∈
/ E3 ),
(32)
where E3 is a set of finite logarithmic measure on (0, +∞). Furthermore, obviously,
R(r)
∼ ln2 ln Sf (r) (r → +∞).
R(r) − r
(33)
Let C3 = C1 + 1, where C1 is the constant from Theorem 1. Using this theorem with
R = R(r) and taking into account (32) and (33), we see that for the random entire function
defined by (2) a. s. for each a ∈ C the inequality (6) holds.
Proof of Corollary 2. Let f be an entire function of form (1), for which
p condition (7) holds,
and let r0 ∈ (0, +∞) be a fixed number such that Sf (r0 ) ≥ max{e, 1 + |c0 |2 }. Put
y(r) =
ln Sf (r)
ln r ln ln Sf (r)
2
(r > r0 ).
(34)
Then, obviously, condition (7) is equivalent to the condition
lim y(r) = +∞.
(35)
r→+∞
First we prove that (35) implies the existence of a set E of upper density 1 on (0, +∞) such
that
lim y(r) = +∞.
(36)
E3r→+∞
There is nothing to prove if the limit λ:= limr→+∞ y(r) is equal to +∞. Let λ < +∞,
and let (λn ) be an arbitrary sequence from the interval (λ, +∞) increasing to +∞. Taking
into account that the function y(r) is continuous on (r0 , +∞) and using (35), it is easy to
justify the existence of sequences (sn ) and (tn ) increasing to +∞ such that r0 < s0 < t0 <
s1 < t1 <S. . . , y(sn ) = 4λn , y(tn ) = λn , and λn ≤ y(r) ≤ 4λn for r ∈ [sn , tn ] and all n ≥ 0.
Put E = ∞
n=0 [sn , tn ]. Then obviously (36) holds. We show that E is a set of upper density 1
on (0, +∞). Indeed, since the function
h(r) =
ln Sf (r)
ln ln Sf (r)
is increasing on (r1 , +∞), we have that
ln2 tn − ln2 sn =
This implies that sn <
h(tn ) h(sn )
3h(sn )
−
>
= 3 ln2 sn
λn
4λn
4λn
(n ≥ n0 ).
√
tn (n ≥ n0 ). Therefore,
Z
dt
dt
tn − sn
≥ lim
= lim
= 1,
n→∞ E∩(s ,t ) tn
n→∞
tn
E∩(0,r) r
n n
Z
lim
r→+∞
(37)
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
47
i. e. the set E has upper density 1 on (0, +∞).
Consider the function lf , introduced in Theorem 2. It is clear that this function is increasing on (r0 , +∞), and therefore, by Theorem 2, for the random entire function defined by (2)
a. s. for each a ∈ C and arbitrary α, β ∈ R, α < β ≤ α + 2π, we obtain
q
β−α
ln Sf (r) + C2 3 ln2 Sf (r)lf (r) ln2 r (r ≥ r4 (ω, a)).
(38)
Nfω (r, α, β, a) ≤
2π
Let E3 be a set of finite logarithmic measure on (0, +∞) for which (32) holds. Then
lf (r) ≤ 2 ln ln Sf (r) for all r ≥ r2 , r ∈
/ E3 , and from (38) we have
q
β−α
ln Sf (r) + C2 3 2 ln2 Sf (r) ln ln Sf (r) ln2 r =
Nfω (r, α, β, a) ≤
2π
s
!
β−α
2
+ C2 3
(r ≥ r5 (ω, a), r ∈
/ E3 ).
(39)
= ln Sf (r)
2π
y(r)
Put E4 = E\E3 . It is clear that the set E4 has upper density 1 on (0, +∞). Using (39)
and (36), a. s. for each a ∈ C and arbitrary α, β ∈ R, α < β ≤ α + 2π, we obtain
β−α
Nfω (r, α, β, a)
≤
.
E4 3r→+∞
ln Sf (r)
2π
lim
(40)
Then, obviously, the validity of relation (4) as E4 3 r → +∞ follows from the equality
Nfω (r, α, β, a) + Nfω (r, β, α + 2π, a) = Nfω (r, a),
(41)
inequality (6) and inequality (40), applied to the angles β and α + 2π instead of the angles
α and β, respectively.
Proof of Corollary 3. Let ρ ∈ (0, +∞), and f be an entire function of the order ρf ≥ ρ and
form (1). It is well known that in the definition of ρf the characteristic Mf (r) can be replaced
with the characteristic Sf (r), i. e.
ln ln Sf (r)
= ρf ≥ ρ.
r→+∞
ln r
lim
We consider the set E = {r > r0 : ln ln Sf (r) > ρ2 ln r} and prove that its upper density on
(0, +∞) is equal to 1. There is nothing to prove if there exists r1 > 0 such that ln ln Sf (r) >
ρ
ln r (r ≥ r1 ). Otherwise, E, as an open set, we can represent in the form of a countable
2
union of intervals. From this union one can choose a sequence of intervals ((sn , tn )) such that
for every n ≥ 0 we have sn < tn < sn+1 , ln ln Sf (sn ) = ρ2 ln sn , ln ln Sf (tn ) = ρ2 ln tn , and
there exists xn ∈ (sn , tn ) such that ln ln Sf (xn ) = 2ρ
ln xn . Then
3
ln tn =
2
2
4
ln ln Sf (tn ) > ln ln Sf (xn ) = ln xn ,
ρ
ρ
3
from which we obtain the relation sn = o(tn ), n → +∞. This relation implies (43), i. e. the
set E has upper density 1 on (0, +∞).
Put E5 = E\E3 , where E3 is a set of finite logarithmic measure on (0, +∞) for which (32)
is satisfied. The set E5 has upper density 1 on (0, +∞) and, according to (39), for the random
48
M. P. MAHOLA, P. V. FILEVYCH
entire function defined by (2) a. s. for each a ∈ C and arbitrary α, β ∈ R, α < β ≤ α + 2π,
we have
r
4
β−α
ln Sf (r) + C2 3 2 ln2 Sf (r) ln ln Sf (r) 2 ln2 ln Sf (r) ≤
Nfω (r, α, β, a) ≤
2π
ρ
β−α
2C2 2
≤
ln Sf (r) + p
ln 3 Sf (r) ln ln Sf (r) (r ≥ r5 (ω, a), r ∈ E5 ).
3
2π
ρ2
Finally, using equality (41) and inequality (6), and putting C4 = 2C2 , we complete the proof
of Corollary 3.
Proof of Corollary 4. Let f be an entire function of finite order and form (1). Corollary 4 is
obvious if f is a polynomial.
Let the function f be transcendental. Then ln r = o(ln Sf (r)) (r → +∞). In addition,
ln ln Sf (r) ≤ 2ρ ln r (r ≥ r6 ). Therefore, using Theorem 1, for the random entire function
defined by (2) a. s. for each a ∈ C we obtain
ln Sf (r) ≤ Nfω (r, a) + C1 ln ln Sf (2r) + ln 2 + h(|a|) (r ≥ r1 (ω)).
This implies that
ln Sf (r)
≤ 1.
r→+∞ Nfω (r, a)
lim
On the other hand, using (21) with fω − a instead of f and (18) with fω instead of f , for
arbitrary ω ∈ Ω and a ∈ C we have
Nfω (r, a) ≤ ln Sf (r) +
1
+ ln+ |a| + ln 2 − ln |cfω (a)|.
2e
From this it follows that
ln Sf (r)
≥ 1.
r→+∞ Nfω (r, a)
lim
Therefore, a. s. for every a ∈ C relation (9) holds.
Proof of Corollary 5. Let f be an entire function of finite order ρ and form (1) such that
relation (10) holds. For the function lf , introduced in Theorem 2, we have lf (r) ≤ 2ρ ln r
(r ≥ r7 ). Therefore, using (38), for the random entire function defined by (2) a. s. for each
a ∈ C and arbitrary α, β ∈ R, α < β ≤ α + 2π, we obtain
β−α
Nfω (r, α, β, a) ≤
ln Sf (r) + C2 ln r
2π
q
3
2ρ ln2 Sf (r) (r ≥ r6 (ω, a)).
This together with (10) implies that
Nfω (r, α, β, a)
β−α
≤
.
r→+∞
ln Sf (r)
2π
lim
Using equality (41) and Corollary 4, we complete the proof of Corollary 5.
THE ANGULAR VALUE DISTRIBUTION OF RANDOM ANALYTIC FUNCTIONS
49
Proof of Corollary 6. Let R ∈ (0, +∞), and let f ∈ H(R) be an analytic function of form (1)
1
(r ∈ (0, R)) and prove that (11)
such that relation (11) holds. Put y(r) = ln Sf (r)/ln R−r
implies the existence of a set E of upper density 1 on (0, R) such that
lim
E3r→+∞
y(r) = +∞.
(42)
Then, obviously, (36) holds. We show that E is a set of upper density 1 on (0, +∞).
Indeed, since the function
ln Sf (r)
h(r) =
ln ln Sf (r)
is increasing on (r1 , +∞), one has that
ln2 tn − ln2 sn =
h(tn ) h(sn )
3h(sn )
−
>
= 3 ln2 sn
λn
4λn
4λn
(n ≥ n0 ).
√
tn (n ≥ n0 ). Therefore,
Z
Z
tn − sn
dt
dt
lim
≥ lim
= lim
= 1,
r→+∞ E∩(0,r) r
n→∞ E∩(s ,t ) tn
n→∞
tn
n n
This implies that sn <
(43)
i. e. the set E has upper density 1 on (0, +∞).
There is nothing to prove if the limit λ:= limr→+∞ y(r) is equal to +∞. Let λ < +∞,
and (λn ) be an arbitrary sequence from the interval (λ, +∞) increasing to +∞. Taking into
account that the function y(r) is continuous on (0, R), and using (11), it is easy to justify
the existence of sequences (sn ) and (tn ) increasing to R such that 0 < s0 < t0 < s1 < t1 <
. . . , y(s
S∞n ) = 2λn , y(tn ) = λn , and λn ≤ y(r) ≤ 2λn for r ∈ [sn , tn ] and all n ≥ 0. Put
E = n=0 [sn , tn ]. Then, obviously, (42) holds. In addition,
ln
ln Sf (tn )
ln Sf (sn )
1
1
=
>
= 2 ln
.
R − tn
λn
λn
R − sn
This implies that R − tn < (R − sn )2 (n ≥ 0). Therefore,
Z
Z
(R − r)dt
(R − tn )dt
R − tn
lim
≥ lim
= lim 1 −
= 1,
n→∞ E∩(s ,t ) (R − t)2
n→∞
r→R E∩(0,r) (R − t)2
R − sn
n n
i. e. the set E has upper density 1 on (0, R). p
Let r0 = min{r ∈ [0, R) : Sf (r) ≥ max{ee , 1 + |c0 |2 }}, and x0 = ln Sf (r0 ). Put
1
R(r) = R − (R − r) exp − 2
(r ∈ [r0 , R)).
ln ln Sf (r)
Applying Lemma 3 for the functions u(r) = ln Sf (r) (r ∈ [r0 , R)) and ϕ(x) = x2 (x ≥ x0 ),
we have
ln Sf (R(r)) < e ln Sf (r) (r ∈ [r0 , R), r ∈
/ E7 ),
(44)
where E7 is a set of finite logarithmic measure on (0, R). Furthermore, obviously,
R(r)
R
∼
ln2 ln Sf (r) (r → R).
R(r) − r
R−r
(45)
50
M. P. MAHOLA, P. V. FILEVYCH
Let C3 = C1 + 1, where C1 is the constant from Theorem 1. Using this theorem with
R = R(r) and taking into account (44) and (45), we see that for random analytic function (2)
a. s. for each a ∈ C the inequality
ln Sf (r) ≤ Nfω (r, a) + C3 ln ln Sf (r) + ln
1
+ h(|a|) (r7 (ω) ≤ r < R)
R−r
(46)
holds.
We consider the function lf , introduced in Theorem 2. Putting C11 = C2 ln2 rR0 , by
Theorem 2 for the random analytic function defined by (2) a. s. for each a ∈ C and arbitrary
α, β ∈ R, α < β ≤ α + 2π, we obtain
q
β−α
Nfω (r, α, β, a) ≤
ln Sf (r) + C11 3 ln2 Sf (r)lf (r) (r8 (ω, a) ≤ r < R).
(47)
2π
Next we note that (44) and (45) implies the inequality
lf (r) ≤ 2(ln ln Sf (r) − ln(R − r)) (r ∈ [r9 , R), r ∈
/ E7 ).
Using this inequality together with (47), we have
s
1
β−α
3
2
ln Sf (r) + C11 2 ln Sf (r) ln ln Sf (r) + ln
=
Nfω (r, α, β, a) ≤
2π
R−r
s
!
2
β−α
2 ln ln Sf (r)
+ C2 3
+
(r10 (ω, a) ≤ r < R, r ∈
/ E7 ).
= ln Sf (r)
2π
ln Sf (r)
y(r)
(48)
Put E6 = E\E7 . It is clear that the set E6 has upper density 1 on (0, R). Using (42) and
(48), a. s. for every a ∈ C and all α, β ∈ R, α < β ≤ α + 2π, we obtain
β−α
Nfω (r, α, β, a)
≤
.
E6 3r→R
ln Sf (r)
2π
lim
(49)
Then, obviously, the validity of the relation (4) as E6 3 r → R follows from equality (41),
inequality (46) and inequality (49), applied to the angles β and α + 2π instead of the angles
α and β, respectively.
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Pidstryhach Institute for Applied Problems of
Mechanics and Mathematics of NASU,
[email protected]
S. Z. Gzhytskyy Lviv National University of
Veterinary Medicine and Biotechnologies,
[email protected]
Received 5.09.2011
Revised 18.01.2012